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1 year ago

15 points!!

Mathematics

2 answers:

Katarina [22]1 year ago

8 0

Answer:

C. 394.2 ft²

Explanation:

surface area of triangular prism: (a+ b + c) × length + base × height

Here given:

a = 9 ft

b = 9 ft

c = 9 ft

base = 9 ft

length = 12 ft

height = 7.8 ft

Applying the formula:

(9+ 9 + 9) × 12 + 9 × 7.8

394.2 ft²

saveliy_v [14]1 year ago

6 0

Answer:

Step-by-step explanation:

9x7.8x1/2x12

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The number of guests per month at a large resort is given in the table below, where f(x) is the number of guests, in hundreds, x

Solnce55 [7]

X = 0 ; y = 10 ; 10 = a(0) + b(0) + c
c = 10

x=2 ; y = 15 ; 15 = a(4) + b(2) + 10 ; 5 = 4a+2b
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y = (1/4)x^2 + 2x + 10 ; 4y = x^2 + 8x + 40

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3 years ago

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I need to see how to work this problem 6-(-7)=

alisha [4.7K]

Hi there!

Here is what the rules says

+ * + = + (Plus times plus = plus)
+*- = - (Plus times minus = minus)
- * - = + (Minus times minus = Plus
- * + = - (minus times plus = minus)

You need to know the rules by heart because you will use it a lot.

So let see how it works with the problem

6 -(-7)
What does the rules say? It says
- * - = plus
So, we gonna use PLUS for the sign
6-(-7) = 6 + 7 = 13

13 is the correct answer.

I hope I helped. Please let me know if you have more questions.

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3 years ago

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Evaluate 35 + 7 -[(12÷4×3)+1]

masha68 [24]

Answer:

Step-by-step explanation:

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3 years ago

How to work it out logically? Question a !!!

Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

$=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha$

Area of region 2 = ar(DHBC)

$=2000\times1500\\\\=3000000m^2=300ha$

Area of region 3 = ar(GFEH)

$(2000+1500)\times 1000\\\\=3500000m^2=350ha$

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

=987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ $ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m$

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

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2 years ago

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Which problem <br> Can we solve with 6x6

Ainat [17]

you can solve option C with 6x6

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2 years ago

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